对数微分

Let's start off with a nice, simple question on derivatives:

As you're probably thinking right now, this function is way too complicated to differentiate directly. (Of course you still can, but that would require lots of paper and time.) To avoid such problems, we use a method called logarithmic differentiation, which uses the ideas of implicit differentiation.

Before we begin, let's remind ourselves of a few rules regarding logarithms:

To recap, these are some of the rules of logarithms that we'll be using:

1. $\ln{x}+\ln{y}=\ln{(xy)}$

2. $\ln{x}-\ln{y}=\ln{\left(\frac{x}{y}\right)}$

3. $\ln{\left(x^a\right)}=a\ln{x}$

Now, let's apply these laws. Suppose we start with a general equation $$y=f(x).$$ We first take the logarithm of both sides to get $$\ln{y}=\ln{f(x)}.$$ Next, we differentiate both sides. Remembering the chain rule and the fact that that $\frac{d}{dx}(\ln{x})=\frac{1}{x}$, we have $$\frac{1}{y}\cdot\frac{dy}{dx}=\frac{d}{dx}(\ln{f(x)}).$$ So, our final expression for the derivative is $$\frac{dy}{dx}=y\cdot\frac{d}{dx}(\ln{f(x)}).$$ How will this be useful, you may ask? Well, let's take the example problem from above, and let $f(x)=(x+13)(x-509)^2(x-1)^3(x+2)^4$. Plugging this in, we get $$\frac{dy}{dx}=f(x)\cdot\frac{d}{dx}\left(\ln{\left[(x+13)(x-509)^2(x-1)^3(x+2)^4\right]}\right).$$ Specifically, let's look at the complicated term on the right hand side: $$\ln{\left[(x+13)(x-509)^2(x-1)^3(x+2)^4\right]}.$$ This is the logarithm of a bunch of products! So, by using the log laws discussed above, we can simplify this as $$\ln{(x+13)}+\ln{(x-509)^2}+\ln{(x-1)^3}+\ln{(x+2)^4}$$ $$=\ln{(x+13)}+2\ln{(x-509)}+3\ln{(x-1)}+4\ln{(x+2)}.$$ So, it follows that $$\small{\frac{dy}{dx}=f(x)\cdot\frac{d}{dx}\left(\ln{(x+13)}+2\ln{(x-509)}+3\ln{(x-1)}+4\ln{(x+2)}\right)}$$ $$\small{=f(x)\cdot\left[\frac{d}{dx} \ln{(x+13)}+2\frac{d}{dx} \ln{(x-509)}+3\frac{d}{dx} \ln{(x-1)}+4\frac{d}{dx} \ln{(x+2)}\right]}$$ $$\small{=(x+13)(x-509)^2(x-1)^3(x+2)^4\cdot\left[\frac{1}{x+13}+\frac{2}{x-509}+\frac{3}{x-1}+\frac{4}{x+2}\right].}$$

This example was still relatively messy, but it required a lot less work than the usual method of differentiation. This shows you how powerful this method can be at times.

Now you try!

We use the method of logarithmic differentiation, and start by taking the logarithm of both sides to get $$\ln{y}=\ln{x^x}=x\ln{x}.$$ Now, we differentiate both sides: $$\frac{1}{y}\cdot\frac{dy}{dx}=\frac{d}{dx}(\ln{y})=\frac{d}{dx}(x\ln{x})=x\frac{d}{dx}(\ln{x})+\ln{x}\frac{d}{dx}(x)=\ln{x}+1.$$ So, it follows that $$\frac{dy}{dx}=y(\ln{x}+1)=x^x(\ln{x}+1).$$

In general, you can see that this method can be very useful for differentiating functions with several products or complicated exponents.

(Videos) Logarithmic Differentiation

by PatrickJMT

Here's a video by patrickJMT showing how to solve problems using logarithmic differentiation. Logarithmic differentiation is different than derivatives of logarithms (something I just learned while writing this). Logarithmic differentiation is a technique where you take the logarithm of both sides in order to simplify the problem. Then you use implicit differentiation to solve for $\frac{dy}{dx}$.

The first problem he looks at is: $$y=\sqrt{x}\cdot e^{x^2}\cdot(x^2+1)^{10}$$ Now you can use the product rule and solve this, but honestly that doesn't sound or look fun. What we can do is take the natural log of both sides. $$\begin{align} \ln y&=\ln[\sqrt{x}\cdot e^{x^2}\cdot(x^2+1)^{10}]\\ \ln y&=\ln(x^{\frac12})+\ln(e^{x^2})+\ln(x^2+1)^{10}\\ \ln y&=\frac12\ln x+x^2\cdot\ln e+10\ln(x^2+1)\\ \ln y&=\frac12\ln x+x^2+10\ln(x^2+1) \end{align}$$ Now we can use implicit differentiation and solve for $\frac{dy}{dx}$. $$\begin{align} \frac1y\cdot\frac{dy}{dx}&=\frac12\cdot\frac1x+2x+10\cdot\frac{1}{x^2+1}\cdot2x\\ \frac1y\cdot\frac{dy}{dx}&=\frac{1}{2x}+2x+\frac{20x}{x^2+1}\\ \frac{dy}{dx}&=y\bigg[\frac{1}{2x}+2x+\frac{20x}{x^2+1}\bigg] \end{align}$$ We want this to be only in terms of $x$, so how do we get rid of that $y$? Well we can substitute in the beginning of the problem, $y=\sqrt{x}\cdot e^{x^2}\cdot(x^2+1)^{10}$. $$\boxed{y'=[\sqrt{x}\cdot e^{x^2}\cdot(x^2+1)^{10}]\left[\frac{1}{2x}+2x+\frac{20x}{x^2+1}\right]}$$

The second problem he looks at is: $$y=(\ln x)^x$$ So the way we solve this is by taking the natural log of both sides and then using implicit differentiation. $$\begin{align} \ln y&=\ln[(\ln x)^x]\\ \ln y&=x\cdot \ln(\ln x) \end{align}$$ So now we can use implicit differentiation to solve for $\frac{dy}{dx}$. $$\begin{align} \color{red}{[\ln y]'}&=\color{red}{[x\cdot \ln(\ln x)]'}\\ \frac1y\cdot\color{red}{(y)'}&=\color{red}{(x)'}\ln(\ln x)+x\color{red}{[\ln(\ln x)]'}\\ \frac1y\frac{dy}{dx}&=\ln(\ln x)+x\cdot\frac{1}{\ln x}\cdot\color{red}{(\ln x)'}\\ \frac1y\frac{dy}{dx}&=\ln(\ln x)+x\cdot\frac{1}{\ln x}\cdot\frac1x\\ \frac1y\frac{dy}{dx}&=\ln(\ln x)+\frac{x}{x\ln x}\\ \frac1y\frac{dy}{dx}&=\ln(\ln x)+\frac{1}{\ln x}\\ \frac{dy}{dx}&=y\left[\ln(\ln x)+\frac{1}{\ln x}\right]\\ \end{align}$$ Now we can use the beginning of the problem to substitute into $y$ in order to have the derivative only contain $x$'s. $$\begin{align} \frac{dy}{dx}&=\color{red}{[(\ln x)^x]}\left[\ln(\ln x)+\frac{1}{\ln x}\right]\\ &=\boxed{(\ln x)^x\left[\ln(\ln x)+\frac{1}{\ln x}\right]} \end{align}$$

Here's another video with another example of logarithmic differentiation. The example being: $$y=\frac{(\sin x)^2(x^3+1)^4}{(x+3)^8}$$ We start by taking the log of both sides. $$\begin{align} \ln y&=\ln\bigg[\frac{(\sin x)^2(x^3+1)^4}{(x+3)^8}\bigg]\\ \ln y&=\ln[(\sin x)^2(x^3+1)^4]-\ln[(x+3)^8]\\ \ln y&=\ln[(\sin x)^2]+\ln[(x^3+1)^4]-\ln[(x+3)^8]\\ \ln y&=2\ln(\sin x)+4\ln(x^3+1)-8\ln(x+3)\\ \end{align}$$ Now we can use implicit differentiation. $$\begin{align} \color{red}{(\ln y)'}&=\color{red}{[2\ln(\sin x)+4\ln(x^3+1)-8\ln(x+3)]'}\\ \frac1y\color{red}{(y)'}&=2\cdot\frac{1}{\sin x}\cdot\color{red}{(\sin x)'}+4\cdot\frac{1}{x^3+1}\cdot\color{red}{(x^3+1)'}+8\cdot\frac{1}{x+3}\\ \frac1y\cdot\frac{dy}{dx}&=\frac{2}{\sin x}\cdot\cos x+\frac{4}{x^3+1}\cdot3x^2+\frac{8}{x+3}\\ \frac1y\cdot\frac{dy}{dx}&=\frac{2\cos x}{\sin x}+\frac{12x^2}{x^3+1}+\frac{8}{x+3}\\ \frac1y\cdot\frac{dy}{dx}&=2\cot x+\frac{12x^2}{x^3+1}+\frac{8}{x+3}\\ \frac{dy}{dx}&=y\bigg[2\cot x+\frac{12x^2}{x^3+1}+\frac{8}{x+3}\bigg]\\ \end{align}$$ Substitute in $y=\frac{(\sin x)^2(x^3+1)^4}{(x+3)^8}$. $$\boxed{y'=\frac{(\sin x)^2(x^3+1)^4}{(x+3)^8}\bigg[2\cot x+\frac{12x^2}{x^3+1}+\frac{8}{x+3}\bigg]}$$

Here's one more video with one final example of logarithmic differentiation. $$y=(x^2)^{\sin x}$$ As always, start by taking the logarithm of both sides. $$\begin{align} \ln y&=\ln[(x^2)^{\sin x}]\\ \ln y&=\sin x\cdot\ln(x^2) \end{align}$$ Now we differentiate both sides. $$\begin{align} \color{red}{[\ln y]'}&=\color{red}{[\sin x\cdot\ln(x^2)]'}\\ \frac1y\cdot\color{red}{(y)'}&=\color{red}{(\sin x)'}\ln(x^2)+\sin x\color{red}{[\ln(x^2)]'}\\ \frac1y\cdot\frac{dy}{dx}&=\cos x\cdot\ln(x^2)+\sin x\cdot\frac{1}{x^2}\cdot\color{red}{(x^2)'}\\ \frac1y\cdot\frac{dy}{dx}&=\cos x\cdot\ln(x^2)+\sin x\cdot\frac{1}{x^2}\cdot2x\\ \frac1y\cdot\frac{dy}{dx}&=\cos x\cdot\ln(x^2)+\frac{2x\sin x}{x^2}\\ \frac1y\cdot\frac{dy}{dx}&=\cos x\cdot\ln(x^2)+\frac{2\sin x}{x}\\ \frac{dy}{dx}&=y\bigg[\cos x\cdot\ln(x^2)+\frac{2\sin x}{x}\bigg] \end{align}$$ Now we substitute in $y=(x^2)^{\sin x}$. $$\boxed{y'=(x^2)^{\sin x}\bigg[\cos x\cdot\ln(x^2)+\frac{2\sin x}{x}\bigg]}$$